Learning Outcomes
- Invert an appropriate matrix to solve a system of linear equations.
Section 4.3 Solving Systems with Matrix Inverses (MX3)
Subsection 4.3.1 Class Activities
Activity 4.3.1.
Consider the following linear system with a unique solution:
\begin{equation*}
\begin{matrix}
3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\
2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\
-x_{1} & & & + & x_{3} & & & = & -1 \\
& - & x_{2} & & & - & 2x_{4} & = & -5 \\
\end{matrix}
\end{equation*}
(a)
Define
\begin{equation*}
T\left(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\right)=
\left[\begin{matrix}\hspace{1em}\unknown\hspace{1em}\\\hspace{1em}\unknown\hspace{1em}\\
\hspace{1em}\unknown\hspace{1em}\\\hspace{1em}\unknown\hspace{1em}\end{matrix}\right]
\end{equation*}
so that \(T(\vec x)=\left[\begin{matrix}-7\\-1\\-1\\-5\end{matrix}\right]\) has the same solution set as this system.
(b)
Define
\begin{equation*}
A=
\left[\begin{matrix}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown
\end{matrix}\right]
\end{equation*}
so that \(A\vec x=\left[\begin{matrix}-7\\-1\\-1\\-5\end{matrix}\right]\) has the same solution set as this system.
(c)
Find
\begin{equation*}
B=
\left[\begin{matrix}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown
\end{matrix}\right]
\end{equation*}
so that \(BA\vec x=\vec x\text{.}\)
(d)
Find \(\vec x=BA\vec x=B\left[\begin{matrix}-7\\-1\\-1\\-5\end{matrix}\right]\) to solve the system.
Remark 4.3.2.
The linear system described by the augmented matrix \([A \mid \vec w]\) has exactly the same solution set as the matrix equation \(A\vec x=\vec w\text{.}\)
Activity 4.3.3.
Let \(A\vec x=\vec w\) describe a linear system. When will this linear system have exactly one solution?
- When \(A\) is invertible.
- When \(A\) is not invertible.
- When \(\RREF A\) has a non-pivot column.
- When \(\RREF A\) has a non-pivot row.
Fact 4.3.4.
When \(A\vec x=\vec w\) has exactly one solution, this solution is given by \(\vec x=A^{-1}\vec w\text{.}\)
Activity 4.3.5.
Consider the vector equation
\begin{equation*}
x_{1} \left[\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right] + x_{2} \left[\begin{array}{c} -2 \\ -3 \\ 3 \end{array}\right] + x_{3} \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] = \left[\begin{array}{c} -3 \\ 5 \\ -1 \end{array}\right]
\end{equation*}
with a unique solution.
(a)
Explain and demonstrate how this problem can be restated using matrix multiplication.
(b)
Use the properties of matrix multiplication to find the unique solution.
Subsection 4.3.2 Videos
Video coming soon to this YouTube playlist 1 .
Exercises 4.3.3 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/exercises/#/bank/MX3/.Subsection 4.3.4 Mathematical Writing Explorations
Exploration 4.3.6.
Use row reduction to find the inverse of the following general matrix. Give conditions on which this inverse exists.
\begin{equation*}
\left[\begin{array}{ccc}1 & b & c \\ d & e & f \\ g & h & i \end{array}\right]
\end{equation*}
Exploration 4.3.7.
Assume that \(H\) is invertible, and that \(HG\) is the zero matrix. Prove that \(G\) must be the zero matrix. Would this still be true if \(H\) were not invertible?Exploration 4.3.8.
If \(H\) is invertible and \(r \in \mathbb{R}\text{,}\) what is the inverse of \(rH\text{?}\)Exploration 4.3.9.
If \(H\) and \(G\) are invertible, is \(H^{-1} + G^{-1} = (H+G)^{-1}\text{?}\)Exploration 4.3.10.
If \(A\) is nonsingular and square, and both \(P\) and \(Q\) are nonsingular, with \(PAQ = I\text{,}\) prove that \(A^{-1} = QP\text{.}\)Subsection 4.3.5 Sample Problem and Solution
Sample problem Example B.1.20.
www.youtube.com/watch?v=kpOK7RhFEiQ&list=PLwXCBkIf7xBMo3zMnD7WVt39rANLlSdmj
